Proposition 6

Let A be a noetherian ring, M a finitely generated A-module and q an ideal of A such that M/qM is of finite length. Let F be a q-good filtration of M, and let $\mathrm{gr}(A) = \bigoplus_{n \geq 0} (q^n/q^{n+1})$ be the graded ring associated with A for the q-adic filtration.

Let $(x_1, ..., x_s)$ be a sequence of elements of A, $(\delta_1, ..., \delta_s)$ a sequence of strictly positive integers such that $x_i \in q^{\delta_i}$ for $1 \leq i \leq s$, and let $\xi_i$ be the class of $x_i$ in $\mathrm{gr}_{\delta_i}(A) = q^{\delta_i}/q^{\delta_i+1}$.

a) Endow the A-module $\overline{M} = M/(x_1M + \cdots + x_sM)$ with the q-good filtration $\overline{F}$ quotient of F. Then one has in $\mathbf{Z}((T))$ the inequality

b) For there to be equality in (22), it is necessary and sufficient that the sequence $(\xi_1, ..., \xi_s)$ of elements of the ring $\mathrm{gr}(A)$ be completely secant for the module $\mathrm{gr}(M) = \bigoplus_{n} (F_n/F_{n+1})$.

In that case, the canonical homomorphism of $\mathrm{gr}(M)/\sum_{i=1}^s \xi_i \cdot \mathrm{gr}(M)$ into $\mathrm{gr}(\overline{M}) = \bigoplus (\overline{F}_n/\overline{F}_{n+1})$ is an isomorphism.

c) Assume the conditions of b) satisfied, and that each of the A-modules $M_i = M/(x_1M + \cdots + x_iM)$ ($0 \leq i < s$) is separated for the q-adic topology¹. Then the sequence $(x_1, ..., x_s)$ is completely secant for the A-module M.

When $s = 1$, one has $\bigcap_n F_n = \bigcap_n q^nM$ and the sequence $\{ \xi_1 \}$ is completely secant for $\mathrm{gr}(M)$ if and only if the homothety of ratio $\xi_1$ in $\mathrm{gr}(M)$ is injective. Prop. 6 then follows at once from lemma 7 applied to the homothety $f = (x_1)_M$ in M.

Assume that one has $s \geq 2$ and argue by induction on s. The induction hypothesis applied to the A-module $M_1 = M/x_1M$ endowed with the filtration G quotient of F, and to the sequence $(x_2, ..., x_s)$ yields the inequality

¹ This occurs in particular if q is contained in the radical of A (III, § 3, no. 3, prop. 6).

there is equality if and only if the sequence $(\xi_2, ..., \xi_s)$ is completely secant for the gr(A)-module $gr(M_1) = \bigoplus_n G_n/G_{n+1}$. Since the elements $\frac{1 - T^{\delta_i}}{1 - T}$ of $\mathbf{Z}((T))$ are positive, the case $s = 1$ already treated and formula (23) yield the inequalities

This proves a).

There can be equality in (22) only if one has simultaneously equality in (23) and the equality

This last relation signifies that $\{\xi_1\}$ is completely secant for gr(M) and implies that the canonical homomorphism of gr(M)/$\xi_1\cdot$gr(M) into gr(M_1) is an isomorphism. In other words, one has equality in (22) if and only if $\{\xi_1\}$ is completely secant for gr(M) and $\{\xi_2, ..., \xi_s\}$ completely secant for gr(M)/$\xi_1\cdot$gr(M). This signifies that $\{\xi_1, ..., \xi_s\}$ is completely secant for gr(M) (A, X, p. 160, Corollary 2). Thus the equivalence of the two conditions of b) has been proved. Suppose them satisfied; then, by the induction hypothesis, gr(M) is identified with $gr(M_1)/\sum_{i=2}^s \xi_i \cdot gr(M_1)$; since moreover gr(M_1) is identified with gr(M)/$\xi_1\cdot$gr(M), the last assertion of b) is therefore satisfied.

Now suppose that $\{\xi_1, ..., \xi_s\}$ is completely secant for gr(M) and M_i separated for the q-adic topology (for $0 \leq i < s$). From the preceding and the induction hypothesis, the sequence $(x_2, ..., x_s)$ is completely secant for M_1; since one has $M_1 = M/x_1M$ and $\{x_1\}$ is completely secant for M, the sequence $(x_1, x_2, ..., x_s)$ is completely secant for M (A, X, p. 160, Theorem 1).